To solve the equation [tex]\( 2x - 5 = -3x^2 \)[/tex], we should first rearrange it into the standard quadratic equation form [tex]\( ax^2 + bx + c = 0 \)[/tex].
Starting with:
[tex]\[ 2x - 5 = -3x^2 \][/tex]
Add [tex]\( 3x^2 \)[/tex] to both sides:
[tex]\[ 3x^2 + 2x - 5 = 0 \][/tex]
Now we have a quadratic equation in the form:
[tex]\[ 3x^2 + 2x - 5 = 0 \][/tex]
We need to find the roots of this quadratic equation. To do this, we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In our equation [tex]\( 3x^2 + 2x - 5 = 0 \)[/tex], the coefficients are:
[tex]\[ a = 3, \quad b = 2, \quad c = -5 \][/tex]
First, calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = 2^2 - 4 \cdot 3 \cdot (-5) \][/tex]
[tex]\[ \Delta = 4 + 60 \][/tex]
[tex]\[ \Delta = 64 \][/tex]
The discriminant is 64, which is a perfect square, so there will be two real and distinct roots. Next, use the quadratic formula to find these roots:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{64}}{2 \cdot 3} \][/tex]
[tex]\[ x = \frac{-2 \pm 8}{6} \][/tex]
This gives us two solutions:
1. Taking the positive square root:
[tex]\[ x = \frac{-2 + 8}{6} = \frac{6}{6} = 1 \][/tex]
2. Taking the negative square root:
[tex]\[ x = \frac{-2 - 8}{6} = \frac{-10}{6} = -\frac{5}{3} \][/tex]
So, the two roots of the equation are:
[tex]\[ x = 1 \][/tex]
[tex]\[ x = -\frac{5}{3} \][/tex]
Therefore, the correct answers are:
C. [tex]\( x = 1 \)[/tex]
D. [tex]\( x = -\frac{5}{3} \)[/tex]
